Question

A marksman hits a target with probability 2/3. Assuming independence for successive firings, find the probabilities of getting
(a) One miss followed by two hits.
(b) Two misses and one hit (in any order).
Round answer to 2 decimal places.

Answer 1

Let p=2/3= probability of success, and q=1/3=probability of failure.

For (a), as these events are independent, you have

q.p.p = 1/3 x (2/3)^2 = 4/27

For (b), you have a binomial probability problem. In n trials, the probability of r successes will be \[^nC_rp^rq^{n-r}=\frac{n!}{r!(n-r)!}p^rq^{n-r}\]You have 3 trials, 1 success. So \[\frac{3!}{1!2!}(\frac{2}{3})(\frac{1}{3})^2=3 \times \frac{2}{27}=\frac{2}{9}\]